Suppose a function fa,b) -> R is continuous and f(r) = 0 for each rational number r in (a,b). Show that f(x) = 0 for each x element (a,b).
So by the definition of continuous every sequence (xn) element dom(f) converges to x0 and limf(xn) = f(x0). Then, xn = 0 must converge to x0. Then, limf(0) = lim0 = 0 = f(x0).
I was wondering if this was correct logic and if there other xn sequences that could be used.

Once again you have the right idea. But you need more detail.
Suppose that $\displaystyle f(x_0 ) \ne 0$. Every real number is the limit of a sequence of rational numbers.
So there is a sequence of rational numbers in (a,b) and $\displaystyle \left( {r_n } \right) \to x_0 $. But $\displaystyle \left( {\forall n} \right)\left[ {f\left( {r_n } \right) = 0} \right]$.
Is that possible if $\displaystyle f(x_0 ) \ne 0$ and $\displaystyle \left( {f\left( {r_n } \right)} \right) \to f\left( {x_0 } \right)$ by continuity?